User john francis - MathOverflow

Failure of smoothing theory for topological 4-manifolds

2009-12-05T23:10:28Z

Smoothing theory fails for topological 4-manifolds, in that a smooth structure on a topological 4-manifold $M$ is not equivalent to a vector bundle structure on the tangent microbundle of $M$. Is there an explicit compact counterexample, i.e., are there two compact smooth 4-manifolds which are homeomorphic, have isomorphic tangent bundles, but are not diffeomorphic? (The uncountably many smooth structures on $\mathbb{R}^4$ should give a noncompact counterexample, since $Top(4)/O(4)$ does not have uncountably many components.)

Addendum to question, added 12/11/09:

I'm also interested in the other type of counterexample, of a nonsmoothable topological 4-manifold whose tangent microbundle does admit a vector bundle structure. Does someone know such an example? Tim Perutz's answer to my first question, below, says that homeomorphic smooth 4-manifolds have isomorphic tangent bundles. If it's not true that all topological 4-manifolds have vector bundle refinements of their tangent microbundle, what is the obstruction in the homotopy of $Top(4)/O(4)$?

Diffeomorphisms vs homeomorphisms of 3-manifolds

2011-12-29T18:08:13Z

For a smooth 3-manifold $M$, is the natural map from the space of diffeomorphisms of $M$ to the space of homeomorphisms of $M$,

$${\sf Diff}(M) \longrightarrow {\sf Top}(M),$$

a weak homotopy equivalence? Equivalently, is the space of smooth structures on a topological 3-manifold contractible? (This is as opposed to just connected, which is the usual statement of Moise's theorem.)

Answer by John Francis for Oriention-Reversing Diffeomorphisms of a Manifold

2010-04-20T19:28:43Z

A large number of manifolds of dimension $4k$ can't admit an orientation-reversing diffeomorphism just because of their cobordism type. That is, if $f: M\rightarrow \overline{M}$ is an orientation preserving diffeomorphism, then the cobordism class $[M^n]$ is a 2-torsion element of the cobordism group of oriented $n$-manifolds: Since $M\sqcup M \cong M\sqcup\overline{M}$ bounds the cylinder $M\times[0,1]$, thus $2[M] = [M\sqcup \overline{M}] = 0 \in \Omega^{\rm SO}_n$. By the Thom-Pontryagin theorem, if $M$ has a nonzero Pontryagin number (which requires that the dimension of $M$ to be a multiple of 4), then $[M]$ is generates a free abelian subgroup of $\Omega^{\rm SO}_n$ and is not a 2-torsion element. Thus, $M$ will not admit an orientation-reversing diffeomorphism.

In particular, this applies if the signature of $M$ is nonzero, since by Hirzebruch's signature theorem the signature is computable in terms of Pontryagin numbers. The previously mentioned examples of $\mathbb{CP}^{2k}$ and $\mathbb{HP}^k$ are special cases of this statement, since both have nonzero signature and hence are do not represent 2-torsion elements of the oriented cobordism group.

Answer by John Francis for How do you compute the space of lifts of an E-infinity map?

2010-02-17T19:33:04Z

If $Y$ and $B$ are grouplike, then the question, of course, immediately reduces to the case of spectra: The map $X\rightarrow B$ factors through the group completion $\Omega B X$ of $X$, so you can then deloop everything to get the corresponding spectra. Once you're in the setting of connective spectra, there are spectral sequences available, the most obvious being the one based on the Postnikov filtration. (This does start with figuring out the $\pi_0$ case, then inductively working through the higher homotopy groups, so affirmative to Q2. I'm not, unfortunately, optimistic about your hope in Q2.)

You can still reduce to the case of spectra with the more modest assumption that the induced map $\pi_0 K \rightarrow \pi_0 \Omega B K$ is an injection, for $K = B, Y$. In this case, you can group complete everything, and then try to solve the problem for the corresponding spectra. An $E_\infty$ lift $X\rightarrow Y$ over $B$ will be the same as a $E_\infty$ lift of the corresponding group completions satisfying an additional $\pi_0$ condition, that the image of $\pi_0 X\rightarrow \pi_0\Omega BX \rightarrow \pi_0 \Omega B Y$ lies in $\pi_0 Y$.

Perhaps you can extend this approach to deal with more general monoids, by writing them as extensions ${\rm Fiber} \rightarrow Y \rightarrow \Omega B Y$. (But you can't be interested in those, can you?)

Answer by John Francis for Model Structure/Homotopy Pushouts in topological monoids?

2010-01-08T02:38:41Z

It's easy to describe the group completion of a pushout, up to homotopy. If $Y\leftarrow X \rightarrow Z$ is the diagram of topological monoids, then the group completion of the pushout is equivalent to $\Omega(BY\amalg_{BX} BZ)$, the based loop space of the pushout of the diagram of pointed spaces $BY \leftarrow BX \rightarrow BZ$.

If your monoids are grouplike to begin with then their pushout is also grouplike, and this gives you an answer.

Answer by John Francis for Diffeomorphism of 3-manifolds

2009-12-15T03:11:00Z

I don't know in general, so I'll just the more obvious cases. For hyperbolic 3-manifolds, this is implied by Mostow's rigidity theorem, which states that a homotopy equivalence of hyperbolic manifolds $n$-manifolds is homotopic to an isometry. It's also true for $S^3$, since both $Diff(S^3)$ and $Aut^h(S^3)$ have two components.

Smooth structures on PL 4-manifolds

2009-12-05T20:16:47Z

Is it known whether $O(4) \to PL(4)$, the map from the orthogonal group to the group of piecewise linear homeomorphisms of $\mathbb{R}^4$, is a homotopy equivalence? By smoothing theory for PL manifolds, this is equivalent to whether the space of smooth structures on a PL 4-manifold is contractible. (I think it's known that this map is at least 4-connected, which shows that the space of smooth structures on any PL 4-manifold is nonempty and connected.)

Comment by John Francis

2011-12-29T19:36:09Z

That's great, thanks.

Comment by John Francis

2011-12-29T19:24:19Z

Here's one definition: let ${\sf Mfld}_3^{\sf sm}$ be the topological groupoid of smooth 3-manifolds with diffeomorphisms, and let ${\sf Mfld}_3$ be the topological one. Let $B$ be the classifying space functor. Then there is a natural map of spaces $$B{\sf Mfld}_3^{\sf sm} \longrightarrow B{\sf Mfld}_3.$$ One definition of the space of smooth structures on $M$ is that it is the homotopy fiber of this map, over the point $\{M\}$ in the base. One reference, for example, is Weiss & Williams' paper, Automorphisms of Manifolds. (This space is naturally a homotopy type , not a homeomorphism type.)

Comment by John Francis

2011-12-29T18:59:22Z

I don't understand why you say that these statements are not equivalent. I think they are. Also, Hatcher and Gabai's results involve the relation of spaces of isometries with spaces of diffeomorphisms, whereas I'm asking about the relation between diffeomorphisms and homeomorphisms. Am I missing something?

Comment by John Francis

2009-12-12T18:13:33Z

Assuming the $o=ks$ claim (that I, unfortunately, don't know how to show at the moment), this seems to me to give that the plumbing of $E_8\oplus E_8$ has a stable tangent bundle. This might be standard, but how do you go from that to the unstable statement? A priori, it's plausible that a microbundle $\tau_M \oplus \mathrm{R}^k$ could admit a vector bundle structure even if $\tau_M$ doesn't.

Comment by John Francis

2009-12-06T23:08:44Z

And do you know an example of distinct smooth, stably parallelizable, compact 4-manifolds that are homeomorphic? Also, I'm skeptical that topological 5-manifolds have unique smoothings; I believe smooth structures on a smoothable topological 5-manifold $M$ are distinguished by an invariant in $H^3(M, \mathbb{Z}/2)$.

Comment by John Francis

2009-12-06T22:49:48Z

Do these papers deal with issue of whether the tangent bundles of these distinct smoothings are isomorphic? At first glance, I don't see that point considered.